3个回答
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(16)
√2[cos(π/4)+isin(π/4)]
=1+i
(17)
|z| = 4
z
=4[cos(π/3)+isin(π/3)]
=4[ (1/2)+(√3/2)i]
=2 + 2√3i
(18)
|6i| =6
arg(6i) = π/2
z
=6[cos(π/2) +isin(π/2) ]
(19)
(x+yi)i +(y+4i) = (x-yi)(1+i)
(x+4)i = (x+y)+(x-y)i
=>
x+y =0 (1)
x+4=x-y (2)
from (2)
y=-4
from (1)
x=4
(x,y)=(4,-4)
√2[cos(π/4)+isin(π/4)]
=1+i
(17)
|z| = 4
z
=4[cos(π/3)+isin(π/3)]
=4[ (1/2)+(√3/2)i]
=2 + 2√3i
(18)
|6i| =6
arg(6i) = π/2
z
=6[cos(π/2) +isin(π/2) ]
(19)
(x+yi)i +(y+4i) = (x-yi)(1+i)
(x+4)i = (x+y)+(x-y)i
=>
x+y =0 (1)
x+4=x-y (2)
from (2)
y=-4
from (1)
x=4
(x,y)=(4,-4)
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2018-08-27 · 知道合伙人教育行家
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